sorry i dont get it ( still learning )
how do i suppose to get tan from it.
do you mean (cos alpha + isin alpha)^2

August 19th 2012, 09:59 PM

richard1234

Re: De Moivre's Theorem

Quote:

Originally Posted by ok59

sorry i dont get it ( still learning )
how do i suppose to get tan from it.
do you mean (cos alpha + isin alpha)^2

Yes, it's also equal to by de Moivre's. I don't know if that helps though. Expanding doesn't yield anything useful.

It could also help to play around with the RHS, . Note that this is a purely imaginary number. You can try to reduce the LHS as or something like that. You'll most probably need your trigonometric identities.

August 19th 2012, 10:08 PM

ok59

Re: De Moivre's Theorem

hmm.. this problem is so freaking weird . I am about to give up ( it has been 2 days).

August 19th 2012, 10:36 PM

richard1234

Re: De Moivre's Theorem

Yeah, it's an interesting problem. I haven't solved it yet...there seem to be several possible ways to the solution though. Just try a bunch of methods, see if one of them works. Maybe another user has a nice, elegant solution that doesn't require a lot of brute force (I don't want to brute force algebra/trig identities right now...). There might even be a geometric solution involving two points on a circle and finding the tangent of the angle in between.

Man ! Thank so very much Deveno.
You made it so simple to understand mate. You can't imagine the time i've spent on this problem.
Thank You again for your help.
Have a nice day !

edit: Thank you to Richard1234 too. I have solved it by your way too. Now I have two solutions.(Cool)

August 20th 2012, 07:45 AM

Soroban

Re: De Moivre's Theorem (Part 1)

Hello, ok59!

Out of curiosity, I brute-forced it . . . whew!

We need the Sum-to-Product Identities:

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August 20th 2012, 01:24 PM

ok59

Re: De Moivre's Theorem (Part 1)

Thanks Mr. Soroban for another method . Now I understand that the trick is to divide and multiply with the conjugate of denominator to get the result. I appreciate the efforts of the members of this board.

September 12th 2012, 11:39 PM

manoj9585

Re: De Moivre's Theorem

De Moivre's Theorem is a relatively simple formula for calculating powers of complex numbers.De Moivre's formula states that for any real number x and any integer n, (cosx + isinx)n = cos(nx) + isin(nx).(r cisθ)^n = r^n cis(nθ) here n is an integer.free algebra word problems

September 13th 2012, 03:04 AM

johnsomeone

Re: [Solved] De Moivre's Theorem

Here's another (equivalent) way: I'll begin with a general calculation that I'll then apply to this specific problem.

Suppose , and (i.e. on the unit circle except the leftmost point).

So let , where is not an odd multiple of .

Then

.

With that in hand, the result follows quickly. Let , . Let . (Note that , so .)

Let . Then , and obviously .

Notice that from , it follows that , so that , and hence so can apply the above derivation with this .

Now consider . Have:

as desired.

Thus the claim is established.

Notice that all the "bad" cases amounted to , because then , so , so .

That would've made , and also would've maded , which is where is undefined.